My point is the same for int as for vector<int>. There is 0 difference in the C++ calling convention between passing a vector<int> and a vector<int>: they both copy an object of the parameter type. Of course, copying a 1000 element vector is much slower than copying a single pointer, but the difference is strictly the size of the type. The copying occurs the same way regardless. This is also the reason foo(char) is less overhead than a foo(char).
Everything (except reference types) is pass-by-value, but of course values can have wildly different sizes.
Also, the problem of accidentally copying large structs is not limited to arguments, the same considerations are important for assignments. Another reason why "pass-by-pointer" shouldn't be presented as some special thing, it's just passing a pointer copy.
Your vector<int*> is a red herring. The distinction I'm making is between passing a (vector<int>)* and a vector<int>, because those two objects have radically different sizes, and the distinction can and does create severe performance issues. And yet, pointers are still different from references: with a reference, you don't even need your object to have a memory address.
HN markup ate my *... Yes, I'm also talking about vector<int> and vector<int>*. They are indeed of radically different sizes, and the consequences of copying one are very different from the consequences of copying the other.
But this doesn't change the fact that they are both passed-by-value when you call a function of that parameter type.
Everything (except reference types) is pass-by-value, but of course values can have wildly different sizes.
Also, the problem of accidentally copying large structs is not limited to arguments, the same considerations are important for assignments. Another reason why "pass-by-pointer" shouldn't be presented as some special thing, it's just passing a pointer copy.